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laiz [17]
4 years ago
5

A cop claims you passed him at more than 60 mph ,27.8 meters per second, 200 meters from a red light that just turned green. you

drive a car that accelerates at a max of 1.5 meters per second. Is the cop correct?
Physics
1 answer:
postnew [5]4 years ago
5 0

Sounds like the situation is modeled like so:

[Light] - - - - - - - - - - [You] => - - -

- - - - - - - - - - - - - - - [Cop] - - - - -

<= = = 200 m = = =>

where your velocity is 27.8 m/s to the right. So the question is asking, is it possible that your car, given that it accelerates at a maximum of 1/5 m/s^2, can reach a speed of 27.8 m/s from rest in the time it takes for it to cover 200 m.

The car's position at time t is

x=\dfrac12\left(1.5\dfrac{\rm m}{\mathrm s^2}\right)t^2

The time it takes to traverse 200 m is

200\,\mathrm m=\left(0.75\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t\approx16.3\,\mathrm s

Your car's velocity at time t, starting from rest, is

v=\left(1.5\dfrac{\rm m}{\mathrm s^2}\right)t

so that after 16.3 s, your car is moving at a speed of

v=\left(1.5\dfrac{\rm m}{\mathrm s^2}\right)\left(16.3\,\mathrm s\right)\approx24.5\dfrac{\rm m}{\rm s}

which is less than the speed the cop claims you were going, so the cop is not correct.

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Suppose we could shrink the earth without changing its mass..?At what fraction of its current radius would the free-fall acceler
Drupady [299]

Answer:

at R/\sqrt{3}

Explanation:

The free-fall acceleration at the surface of Earth is given by

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

The formula can be rewritten as

R=\sqrt{\frac{GM}{g}} (1)

We want to shrink the Earth at a radius R' such that the acceleration of gravity becomes 3 times the present value, so

g' = 3g

Keeping the mass constant, M, and substituting into the equation, we have

3g=\frac{GM}{R'^2}

R'=\sqrt{\frac{GM}{3g}}=\frac{1}{\sqrt{3}}\sqrt{\frac{GM}{g}}=\frac{R}{\sqrt{3}}

5 0
3 years ago
Read 2 more answers
In some applications of ultrasound, such as its use on cranial tissues, large reflections from the surrounding bones can produce
worty [1.4K]

Answer:

b. 0.75 mm

Explanation:

The distance between antinodes d is half the wavelength \lambda. We can obtain the wavelength with the formula v=\lambda f, where f is the frequency given (f=1MHz=1\times10^6Hz) and v is the speed of sound in body tissues (v=1540m/s), so putting all together we have:

d=\frac{\lambda}{2}=\frac{v}{2f}=\frac{1540m/s}{2(1\times10^6Hz)}=0.00077m=0.77mm

which is very close to the 0.75mm option.

4 0
4 years ago
Frequency, period and wavelength<br> 11th grade high school physics
zzz [600]

Answer:0.38

Explanation:

the formula is f = c / λ

so f= 2.5/6.5

and that equals 0.38 46 and so on so i just rounded it

6 0
3 years ago
Violet light of wavelength 427 nm ejects electrons with a maximum kinetic energy of 0.684 eV from a certain metal. What is the w
frutty [35]

Answer:

The work function of the metal is 2.226 eV.

Explanation:

Given;

wavelength of the violet light, λ = 427 nm = 427 x 10⁻⁹ m

maximum kinetic energy, K.E = 0.684 eV

The energy of the incident light is calculated as;

E = hf = \frac{hc}{\lambda} = \frac{6.626 \ \times \ 10^{-34} \ \times\ 3\ \times \ 10^8 }{427 \ \times \ 10^{-9}} = 4.655 \ \times \ 10^{-19} \ J\\\\1 \ eV = 1.6 \ \times \ 10^{-19} \ J\\\\E =( \frac{4.655 \ \times \ 10^{-19} \ J }{1.6 \ \times \ 10^{-19} \ J} ) \ eV\\\\E = 2.91 \ eV

Apply Einstein's photoelectric equation;

E = Ф + K.E

where;

Ф is the work function of the metal

Ф  = E - K.E

Ф  = 2.91 eV - 0.684 eV

Ф  = 2.226 eV.

Therefore, the work function of the metal is 2.226 eV.

5 0
3 years ago
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